Question

If a body is taken above the surface of the Earth, it loses its weight by 20% at a height of \( h = \frac{\sqrt{5}}{2} R \), where \( R \) is the radius of the Earth, then:

• A. \( \frac{\sqrt{5}}{2} R \)
• B. \( \left( \frac{\sqrt{5}}{2} - 3 \right) R \)
• C. \( \left( \frac{\sqrt{5}}{2} - 1 \right) R \)
• D. \( \left( \frac{\sqrt{5}}{2} - 2 \right) R \)
• E. \( \left( 1 + \frac{\sqrt{5}}{2} \right) R \)

Hint:

The weight of a body decreases with the square of the distance from the center of the Earth. Use the formula \( W = W_0 \left( \frac{R}{R + h} \right)^2 \) to calculate the change in weight with height.

Answer 1

The Correct Option is C

The weight of a body at a height \( h \) above the Earth's surface is given by the formula: \[ W = W_0 \left( \frac{R}{R + h} \right)^2 \] where: - \( W_0 \) is the weight at the Earth's surface,
- \( R \) is the radius of the Earth,
- \( h \) is the height above the Earth's surface.
The weight decreases by 20% at height \( h \), so the new weight \( W \) is 80% of the original weight: \[ W = 0.8 W_0 \] Substituting into the formula for \( W \): \[ 0.8 W_0 = W_0 \left( \frac{R}{R + h} \right)^2 \] Canceling \( W_0 \) from both sides: \[ 0.8 = \left( \frac{R}{R + h} \right)^2 \] Taking the square root of both sides: \[ \sqrt{0.8} = \frac{R}{R + h} \] Simplifying: \[ \frac{R}{R + h} = \frac{\sqrt{5}}{2} \] Now, solving for \( h \): \[ R + h = \frac{2R}{\sqrt{5}} \] \[ h = \frac{2R}{\sqrt{5}} - R \] \[ h = R \left( \frac{2}{\sqrt{5}} - 1 \right) \] Simplifying further: \[ h = R \left( \frac{\sqrt{5}}{2} - 1 \right) \] Thus, the height at which the body loses 20% of its weight is: \[ \boxed{\left( \frac{\sqrt{5}}{2} - 1 \right) R} \]